= Solution
For monatomic species, there are no molecular rotational or vibrational factors. Assume excited electronic states are negligible, so the internal partition sums are the ground-state degeneracies. A free <electron> has two <spin> states, giving
$$
f_e=2\left(\frac{2\pi m_ek_{\rm B}T}{h^2}\right)^{3/2},\qquad
f_{\rm M}=g_{\rm M}\left(\frac{2\pi m_{\rm M}k_{\rm B}T}{h^2}\right)^{3/2},
$$
with the analogous expression for the ion. Since losing one <electron> changes the atomic mass negligibly, $(m_{\rm M^+}/m_{\rm M})^{3/2}\simeq1$. Hence
$$
\boxed{K_c=\left(\frac{2\pi m_ek_{\rm B}T}{h^2}\right)^{3/2}
\frac1{c^\circ}\frac{2g_{\rm M^+}}{g_{\rm M}}e^{-E_I/k_{\rm B}T}.}
$$
This is the ground-state form of the <Saha ionization equation>. Its assumptions include a dilute, nondegenerate ideal mixture, negligible interparticle interactions, thermal and <chemical equilibrium>, no additional ionization stages, and consistent treatment of internal degeneracies. Nuclear-spin factors may be omitted when they cancel between the atom and ion; they must not be counted on only one side.
Back to article page
